Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 11x}{x - 1} = \dfrac{-23x - 32}{x - 1}$
Solution: Multiply both sides by $x - 1$ $ \dfrac{x^2 - 11x}{x - 1} (x - 1) = \dfrac{-23x - 32}{x - 1} (x - 1)$ $ x^2 - 11x = -23x - 32$ Subtract $-23x - 32$ from both sides: $ x^2 - 11x - (-23x - 32) = -23x - 32 - (-23x - 32)$ $ x^2 - 11x + 23x + 32 = 0$ $ x^2 + 12x + 32 = 0$ Factor the expression: $ (x + 8)(x + 4) = 0$ Therefore $x = -8$ or $x = -4$ The original expression is defined at $x = -8$ and $x = -4$, so there are no extraneous solutions.